6th grade math help me plzzzzz

Solve by Elimination: <br> 4x - 5y = 14<br> -4x + y = 6

Lana71 [14]

Answer:x = 1 , y = − 2

Step-by-step explanation:

5 0

2 years ago

Festus bought a computer at $736.34, Apple iPad at $249.99, and a printer at 381.23. The sales tax rate is %6. What is the total

tatuchka [14]

The total purchase price including sales tax will be $1449.6136

<em><u>Explanation</u></em>

Festus bought a computer at $736.34, Apple iPad at $249.99 and a printer at $381.23

So, the <u>total price of all three items</u> $=(736.34+249.99+381.23) dollar = 1367.56$ dollar.

The sales tax rate is 6%. So, <u>the amount of sales tax</u> $=(1367.56*0.06) dollar= 82.0536$ dollar.

Thus, the total purchase price including sales tax will be: (1367.56 + 82.0536) dollar = 1449.6136 dollar

6 0

3 years ago

Simplify<br> (2x-³y⁴)³(x³+y⁰)÷(4xy-²)³

elena-14-01-66 [18.8K]

Answer:

Step-by-step explanation:

-4x³(-2x³)

8x⁶

(6⁵)/(6⁴)

6

(3x³)³

27x⁹

(9¹²)/(9⁸)

9⁴

(x²)(x³)

x⁵

(x⁴)/(x²)

x²

-x(-x)(x)

x³

(x³y²)/(x³y⁴)

(1)/(y²)

-2x(x²)(-3x)

6x⁴

(9x⁷)/(3x⁶)

3x

3x²(x²)(-6)

-18x⁴

x/(x³)

1/(x²)

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(x⁴)(x³)

x⁷

(12x⁵)/(36x)

x⁴/3

(x⁴y³)/(x⁴y)

y²

(-2x³)(-4x²)

8x⁵

(-3x²)³

-27x⁶

(x³y⁵)/(xy²)

x²y³

2x³(10x)³

2000x⁶

(x⁷y²)/(x⁴y²)

x³

x⁻⁴

1/x⁴

(-21x⁵y²)/(7x⁴y⁵)

-(3x)/y³

3x⁻³

3/x³

(32x³y²z⁵)/(-8xyz²)

-4x²yz³

...

x⁴y⁴

(4x⁷/7y²)²

(16x¹⁴)/(49y⁴)

...

8x³

4⁻⁴

1/(4⁴) or 1/256

...

a⁵

8⁻²

1/(8²) or 1/64

...

2c⁷

x⁻²

1/x²

...

9x²

x⁻³

1/x³

...

a⁴

x⁻⁴

1/x⁴

...

4c⁵

x²y⁻³

(x²)/(y³)

When multiplying monomials with the same base, ___________ the exponents

add

x³y⁻²

(x³)/(y²)

When dividing monomials with the same base, ______________ the exponents

subtract

x²y³z⁻⁴

(x²y³)/(z⁴)

When raising a power to a power, ________________ the exponents

multiply

x²y⁻³z⁴

(x²z⁴)/(y³)

x⁻²y⁻³

1/(x²y³)

A base raised to a zero exponent equals________________

one

(2m²)(2m³)

4m⁵

(x/y)⁻¹

y/x

4r⁻³(2r²)

8÷r

(x²/y²)⁻¹

y²/x²

2x³y⁻³(2x⁻¹y³)

4x²

(x³/y³)⁻¹

y³/x³

2y²(3x)

6y²x

(x/y)⁻²

y²/x²

4a³b²(3a⁻⁴b⁻³)

12÷(ab)

(x²/y²)⁻³

y⁶/x⁶

4r⁰

4

(2/3)⁻²

9/4

(4xy)⁻¹

1÷(4xy)

(4/5)⁻²

25/16

(3/4)⁻²

16/9

(3/4)⁻¹

4/3

(x³/x⁻⁶)

x⁹

(x²/x⁻⁵)

x⁷

x⁰

1

y⁰

1

x²y⁰

x²

x²y³z⁰

x²y³

y⁰

1

100⁰

1

(-99)⁰

1

x⁰(x⁴)(x⁻⁶)

1/x²

(x⁻³)/(x⁴)

1/x⁷

(x⁻⁴)/(x⁵)

1/x⁹

(4x³/2x⁵)⁰

1

(x⁻⁵y⁴)/(z⁻²)

(y⁴z²)/x⁵

(15x⁶y⁻⁹)/(5xy⁻¹¹)

3x⁵y²

(48x⁶y⁷z⁵)/(-6xy⁵z⁶)

-(8x⁵y²)/z

Is it a monomial? 11

Yes; 11 is a real number and an example of a constant.

Is it a monomial? a - b

No; this is the difference, not the product, of two variables.

Is it a monomial? p²/r²

No; this is the quotient, not the product, of two variables.

Is it a monomial? y

Yes; single variables are monomials.

Is it a monomial? j³k

Yes; this is the product of two variables.

Is it a monomial? 2a + 3b

No; this is the sum of two monomials.

Simplify x²(x³)(x⁶)

x¹¹

Simplify x(x²)(x⁷)

x¹⁰

Simplify (y²z)(yz²)

y³z³

Simplify (y²z²)(y³z)

y⁵z³

Simplify (a²b⁴)(a²b⁴)

a⁴b⁸

Simplify (ab²)(a³b²)

a⁴b⁴

Simplify (2x²)(3x⁵)

6x⁷

Simplify (5x⁷)(4x²)

20x⁹

Simplify (4xy³)(3x³y⁵)

12x⁴y⁸

Simplify (7x⁵y²)(x²y³)

7x⁷y⁵

Simplify (-5x³)(3x⁸)

-15x¹¹

Simplify (-2x⁴y)(-4xy)

8x⁵y²

Simplify (10²)³

10⁶ or 1,000,000

Simplify (x³)¹²

x³⁶

Simplify (-6x)²

36x²

(-3x)³

-27x³

(3xy²)²

9x²y⁴

(2x³y⁴)²

4x⁶y⁸

Find the area of a rectangle if the length is x² and the width is x⁵.

x⁷

Find the area of a square if the side length is xy.

x²y²

Find the area of a triangle with base 9x³ and height 4x.

18x⁴

Simplify (-5x²y)(3x⁴)

-15x⁶y

Simplify (2ab²c²)(4a³b²c²)

8a⁴b⁴c⁴

Simplify (3xy⁴)(-2x²)

-6x³y⁴

(4x³y)(-2x⁵)

-8x⁸y

(-15xy⁴)(-xy³)

15x²y⁷

(-xy)³(xz)

-x⁴y³z

(-4x²y)²(-½xy²)

-8x⁵y⁴

(0.2x²y³)²

0.04x⁴y⁶

(½xy³)²

¼x²y⁶

(0.4x³)³

0.064x⁹

[(x²)²]²

x⁸

[(x²)³]⁴

x²⁴

Find the area of a rectangle whose length is 6x²y⁴ and width is 3xy²

18x³y⁶

Find the area of a triangle whose base is 4x²y and height is 6xy³

12x³y⁴

Find the volume of a cube whose side length is 3x².

27x⁶

Find the volume of a rectangular prism whose side lengths are x³y, xy³, and y.

x⁴y⁵

5 0

2 years ago

Alan buys 5.3 ounces of sour patch kids candy. After sharing with his friends he returns to buy an additional 6.75 ounces. How m

Natasha_Volkova [10]

Answer:

Alan buys 12.05 ounces of sour patch kids candy in all.

Step-by-step explanation:

We are given that Alan buys 5.3 ounces of sour patch kids candy. After sharing with his friends he returns to buy an additional 6.75 ounces.

And we have to find the total ounces he buys in all.

As we know that for finding the total quantity or amount, we will use addition for calculating it.

Firstly, Alan buys = 5.3 ounces of sour patch kids candy

Additional ounces of sour patch kids candy Alan buys = 6.75

So, the total ounces of sour patch kids candy he buys in all = 5.3 + 6.75

= 12.05 ounces

Hence, he buys 12.05 ounces in all.

7 0

2 years ago

Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.

5 0

2 years ago